Change of State Latent Heat

8 MCQs2 revision cards9-step worked example

Source: NCERT Properties of Bulk MatterPYQ coverage: NEET 2020, 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026

Lesson

A common confusion in NEET problems on change of state: treating latent heat as "extra temperature" instead of recognising that temperature stays constant during a phase transition. The heat you supply at the melting or boiling point does not raise temperature — it breaks intermolecular bonds.

What latent heat means. When a substance reaches its melting point (or boiling point), additional heat goes entirely into changing the phase. The formula is Q = mL, where L is the specific latent heat — either L_fusion (solid → liquid) or L_vaporisation (liquid → gas). For water: L_fusion ≈ 3.34 × 10⁵ J/kg; L_vaporisation ≈ 2.26 × 10⁶ J/kg (NCERT Class 11 Physics, Chapter 11, page 7).

The flat region on the heating curve. Plot temperature vs heat supplied for ice → water → steam. You get two flat plateaus — one at 0 °C (fusion) and one at 100 °C (vaporisation). During each plateau, Q = mL applies; between plateaus, Q = mcΔT applies.

Where NEET problems test you. A typical question gives you a mass of ice at some sub-zero temperature and asks for the total heat to convert it fully to steam at 100 °C. You must chain three or more stages: (1) heat ice from initial temperature to 0 °C using Q = mcΔT with c_ice, (2) melt ice at 0 °C using Q = mL_fusion, (3) heat water from 0 °C to 100 °C using Q = mcΔT with c_water, (4) vaporise water using Q = mL_vaporisation. Missing any stage — or applying the wrong specific heat for the wrong phase — costs you the mark.

Watch out: L_vaporisation is roughly 6.8 times L_fusion for water. If your answer for total heat is dominated by the fusion step rather than the vaporisation step, re-check — you may have swapped the two values.

Practice MCQs

Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.

MCQ 1Easy RecallPractice

During the melting of ice at 0 °C and 1 atm pressure, the temperature of the ice-water mixture:

MCQ 2Easy RecallPractice

The SI unit of specific latent heat is:

MCQ 3Easy RecallPractice

For water at standard pressure, the specific latent heat of vaporisation is approximately:

MCQ 4Direct ApplicationPractice

How much heat is required to melt 0.50 kg of ice at 0 °C completely into water at 0 °C? (L_fusion = 3.34 × 10⁵ J/kg)

MCQ 5Direct ApplicationPractice

200 g of ice at 0 °C is mixed with 200 g of water at 80 °C in a thermally insulated container. What is the final temperature of the mixture? (L_fusion = 3.34 × 10⁵ J/kg, c_water = 4.18 × 10³ J/(kg·K))

MCQ 6Direct ApplicationPractice

On a heating curve (temperature vs heat supplied) for a pure substance, the slope of the curve during a phase change is:

MCQ 7CalculationPractice

Calculate the total heat required to convert 100 g of ice at −10 °C to steam at 100 °C. (c_ice = 2.09 × 10³ J/(kg·K), L_fusion = 3.34 × 10⁵ J/kg, c_water = 4.18 × 10³ J/(kg·K), L_vaporisation = 2.26 × 10⁶ J/kg)

MCQ 8CalculationPractice

A 50 g copper calorimeter (c_copper = 390 J/(kg·K)) contains 100 g of water at 30 °C. A 20 g piece of ice at 0 °C is dropped in. Find the final equilibrium temperature. (L_fusion = 3.34 × 10⁵ J/kg, c_water = 4.18 × 10³ J/(kg·K))

Quick recall before you leave

Worked Example

1
Given
- Mass: m = 50 g = 0.050 kg - Initial temperature: T_i = −20 °C - Final temperature: T_f = 40 °C - c_ice = 2.09 × 10³ J/(kg·K) - L_fusion = 3.34 × 10⁵ J/kg - c_water = 4.18 × 10³ J/(kg·K)
2
Required
Total heat Q_total to go from ice at −20 °C to water at 40 °C.
3
Concept
Three stages: (1) heat ice from −20 °C to 0 °C, (2) melt ice at 0 °C, (3) heat water from 0 °C to 40 °C. Each stage uses a different formula or different value of c.
4
Formulas
- Stage 1: Q₁ = m × c_ice × ΔT₁ - Stage 2: Q₂ = m × L_fusion - Stage 3: Q₃ = m × c_water × ΔT₃
5
Substitution
- Q₁ = 0.050 × 2090 × 20 = ? - Q₂ = 0.050 × 3.34 × 10⁵ = ? - Q₃ = 0.050 × 4180 × 40 = ?
6
Calculation
- Q₁ = 0.050 × 2090 × 20 = 2,090 J - Q₂ = 0.050 × 334,000 = 16,700 J - Q₃ = 0.050 × 4180 × 40 = 8,360 J - Q_total = 2,090 + 16,700 + 8,360 = 27,150 J Note on exact values: the temperature intervals (20 K and 40 K) are exact by problem definition and do not limit significant figures.
7
Final answer
Q_total ≈ 2.72 × 10⁴ J The dominant contribution is the melting step (16,700 J, about 61% of the total), followed by heating water (8,360 J, about 31%), then heating ice (2,090 J, about 8%). If a problem involves vaporisation as well, expect the vaporisation term to dominate overwhelmingly.
8
Common trap
Forgetting to use c_ice for the sub-zero heating stage and instead using c_water throughout. Since c_ice ≈ 0.50 × c_water, this error overestimates Q₁ by a factor of 2. Always check: which phase is the substance in during each stage?
9
Similar NEET-style question
A 200 g block of ice at −15 °C is placed in an insulated container with 300 g of water at 50 °C. Determine the final equilibrium temperature and state of the mixture. (Same constants as above.) *Approach: calculate heat available from water cooling to 0 °C, compare with heat needed to warm ice to 0 °C + melt it. If surplus heat remains, all ice melts and final T > 0 °C. If deficit, some ice remains at 0 °C.* ---

Before solving, remember these

Formula

Latent heat

Q = m L during phase transition at constant T. L_fusion (water): 3.34 × 10⁵ J/kg. L_vaporisation (water): 2.26 × 10⁶ J/kg. Latent heat is absorbed/released without temperature change.

-- NCERT, p. 7

Formulas

12 formulas — click to collapse

Bernoulli's equation

P + \frac{1}{2} ρ v^{2} + ρ g h = const

Conservation of energy along a streamline of incompressible non-viscous flow.

Symbol	Quantity	SI Unit
P	pressure	Pa
rho	density	kg/m^3
v	speed	m/s
g	gravity	m/s^2
h	height	m

Valid when

Steady, non-viscous, incompressible flow
Along a single streamline
No work added/removed

Bulk modulus

K = - V \frac{d P}{d V}

Resistance of a material to uniform compression. Inverse: compressibility.

Symbol	Quantity	SI Unit
K	bulk modulus	Pa
V	volume	m^3
P	pressure	Pa

Valid when

Isotropic compression
Within elastic regime

Capillary rise/depression

h = \frac{2 T c o s θ}{ρ g r}

Height a liquid rises (or falls) in a capillary tube. cos(theta) > 0: rises (wetting); < 0: depresses.

Symbol	Quantity	SI Unit
h	capillary height	m
T	surface tension	N/m
theta	contact angle	rad
rho	density	kg/m^3
r	tube radius	m

Valid when

Narrow tube (capillary regime)
Constant theta

Pressure in static fluid

P = P_{0} + ρ g h

Pressure at depth h below free surface of fluid of density rho.

Symbol	Quantity	SI Unit
P	total pressure	Pa
P0	atmospheric/surface pressure	Pa
rho	density	kg/m^3
h	depth	m

Valid when

Static fluid (no flow)
Constant g
Constant rho (incompressible)

Latent heat

Q = m L

Heat absorbed/released during phase change at constant T. L_fusion or L_vaporisation.

Symbol	Quantity	SI Unit
Q	heat	J
m	mass	kg
L	latent heat	J/kg

Valid when

Phase transition (constant T during)
All mass m undergoes the transition

Specific heat / heat capacity

Q = m c Δ T

Heat required to raise mass m by temperature Delta_T. Specific heat c is material property.

Symbol	Quantity	SI Unit
Q	heat	J
m	mass	kg
c	specific heat	J/kg/K
Delta_T	temp change	K

Valid when

No phase change during heating
c approximately constant in temp range

Stefan-Boltzmann radiation law

P = σ ε A T^{4}

Radiation power from a body. Black body epsilon=1; net to surroundings P = sigma*epsilon*A*(T^4 - T_s^4).

Symbol	Quantity	SI Unit
sigma	Stefan-Boltzmann = 5.67e-8	W/m^2/K^4
epsilon	emissivity (0-1)	-
A	surface area	m^2
T	absolute temp	K

Valid when

Body in radiative equilibrium
T in kelvins

Stokes' law (viscous drag on sphere)

F = 6 π η r v

Drag force on a sphere of radius r moving with velocity v through viscous fluid (low Reynolds number).

Symbol	Quantity	SI Unit
F	drag force	N
eta	viscosity	Pa*s
r	sphere radius	m
v	velocity	m/s

Valid when

Smooth, slow flow (low Reynolds number)
Spherical body
Newtonian fluid

Excess pressure inside drop/bubble

Δ P_{drop} = \frac{2 T}{r}; Δ P_{bubble} = \frac{4 T}{r}

Excess internal pressure due to surface tension. Bubble has 2 surfaces, hence factor 4.

Symbol	Quantity	SI Unit
Delta_P	excess pressure	Pa
T	surface tension	N/m
r	radius	m

Valid when

Spherical drop or bubble
Constant T (one fluid pair)

Terminal velocity of sphere in viscous fluid

v_{t} = \frac{2 r ^{2} g ( ρ _{s} - ρ _{f} )}{9 η}

Constant velocity reached when net force is zero (gravity balanced by buoyancy + viscous drag).

Symbol	Quantity	SI Unit
v_t	terminal velocity	m/s
r	sphere radius	m
rho_s	sphere density	kg/m^3
rho_f	fluid density	kg/m^3
eta	viscosity	Pa*s

Valid when

Steady state (net force zero)
Stokes regime applicable

Thermal expansion (linear/area/volume)

\frac{Δ L}{L} = α Δ T; \frac{Δ A}{A} = 2 α Δ T; \frac{Δ V}{V} = β Δ T

Fractional change in length, area, volume per degree temperature change.

Symbol	Quantity	SI Unit
alpha	linear coefficient	1/K
beta	volume coefficient	1/K
Delta_T	temperature change	K

Valid when

Isotropic material
Modest temperature range (alpha ~ constant)

Young's modulus

Y = \frac{F L}{A Δ L}

Ratio of longitudinal stress to longitudinal strain in a stretched wire/rod within elastic limit.

Symbol	Quantity	SI Unit
Y	Young's modulus	Pa
F	applied force	N
A	cross-section area	m^2
L	original length	m
Delta_L	extension	m

Valid when

Within elastic limit (Hooke's law region)
Uniform cross-section
Force along length

Exam Traps & Common Mistakes

These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.

5 items — click to collapse

Trap

Soap bubble vs drop excess pressure

Category: Similar Terms

Student uses 2T/r for soap bubble (drop formula). Bubble has 2 surfaces → 4T/r.

When it triggers

Question mentions soap bubble OR liquid drop OR air bubble in liquid.

How to avoid

Drop in air: 1 surface → 2T/r. Soap bubble in air: 2 surfaces → 4T/r. Air bubble in liquid: 1 surface → 2T/r.

Trap

Young's vs bulk modulus confusion

Category: Similar Terms

Student uses Y formula when problem is about volumetric compression (use K) or vice versa.

When it triggers

Problem describes longitudinal stretching (use Y), volumetric pressure (use K), or shear (use G).

How to avoid

Y: longitudinal stress/strain. K: volumetric. G: shear. Match modulus to deformation type.

Mistake

Uses 2T/r excess-pressure formula for soap bubble. Bubble has 2 surfaces → 4T/r.

Root cause: concept gap

Correction

Drop in air: 1 surface → ΔP = 2T/r. Soap bubble: 2 surfaces (inner + outer) → ΔP = 4T/r. Air bubble inside liquid: 1 surface → 2T/r.

Mistake

Treats v_terminal as linear in r instead of r².

Root cause: formula misuse

Correction

v_t ∝ r² (because Stokes drag ∝ r v, gravity ∝ r³ - r³ = r³_diff). Doubling radius quadruples terminal velocity, not doubles.

Mistake

Uses Young's modulus formula for volumetric compression problem (or vice versa).

Root cause: formula misuse

Correction

Y for longitudinal stretch (FL/A·ΔL); K for volumetric compression (-V·dP/dV); G for shear. Match modulus type to deformation type before computing.

Past Year Questions

15 questions from NEET 2020, 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys. — click to collapse

NEET 2025

A balloon is made of a material of surface tension S and its inflation outlet (from where gas is filled in it) has small area A. It is filled with a gas of density ρ and takes a spherical shape of radius R. When the gas is allowed to flow freely out of it, its radius r changes from R to 0 (zero) in time T. If the speed v(r) of gas coming out of the balloon depends on r as ra and T ∝ Sα Aβ ργ Rδ then 1 1 1 1 7 1 1 3

1a= ,α= ,β=− ,γ= ,δ=

2a= ,α= ,β=−1,γ=+1,δ= 2 2 2 2 2 2 2 2 1 1 1 5 1 1 1 7

3a=− ,α=– ,β=−1,γ=− ,δ=

4a=− ,α=− ,β=−1,γ= ,δ= 2 2 2 2 2 2 2 2

NTA Answer: Option 4(final)

NEET 2025

Consider a water tank shown in the figure. It has one wall at x = L and can be taken to be very wide in the z direction. When filled with a liquid of surface tension S and density ρ, the liquid surface makes angle θ 0 (θ 0 << 1) with the x-axis at x = L. If y(x) is the height of the surface then the equation for y(x) is: dy (takeθ(x)=sinθ(x)=tanθ(x)= ,g is the acceleration due to gravity) dx dy ρg d2y ρg

1= x

2= x dx S dx2 S d2y ρg d2y ρg

3= y

4= dx2 S dx2 S

NTA Answer: Option 3(final)

NEET 2024Revised key

The maximum elongation of a steel wire of 1 m length if the elastic limit of steel and its Young’s modulus, respectively, are 8 × 108 N m–2 and 2 × 1011 N m–2, is:

14 mm

20.4 mm

340 mm

48 mm

NTA Answer: Option 1(revised_final)

NEET 2024Revised key

A thin flat circular disc of radius 4.5 cm is placed gently over the surface of water. If surface tension of water is 0.07 Nm–1, then the excess force required to take it away from the surface is

119.8 mN

2198 N

31.98 mN

499 N

NTA Answer: Option 1(revised_final)

NEET 2024Revised key

A metallic bar of Young’s modulus, 0.5 × 1011 N m–2 and coefficient of linear thermal expansion 10–5 °C–1, length 1 m and area of cross-section 10–3 m2 is heated from 0°C to 100°C without expansion or bending. The compressive force developed in it is :

15 × 103 N

250 × 103 N

3100 × 103 N

42 × 103 N

NTA Answer: Option 2(revised_final)

NEET 2023

The amount of energy required to form a soap bubble of radius 2 cm from a soap solution is nearly (surface tension of soap solution = 0.03 N m–1)

15.06 × 10–4 J

23.01 × 10–4 J

350.1 × 10–4 J

430.16 × 10–4 J

NTA Answer: Option 2(final)

NEET 2023

The venturi-meter works on

1Bernoulli’s principle

2The principle of parallel axes

3The principle of perpendicular axes

4Huygen’s principle

NTA Answer: Option 1(final)

NEET 2023

Let a wire be suspended from the ceiling (rigid support) and stretched by a weight W attached at its free end. The longitudinal stress at any point of cross-sectional area A of the wire is

1W/A

2W/2A

3Zero

42W/A

NTA Answer: Option 1(final)

NEET 2022

If a soap bubble expands, the pressure inside the bubble

1Is equal to the atmospheric pressure

2Decreases

3Increases

4Remains the same

NTA Answer: Option 2(final)

NEET 2022

A spherical ball is dropped in a long column of a highly viscous liquid. The curve in the graph shown, which represents the speed of the ball (v) as a function of time (t) is

NTA Answer: Option 3(final)

NEET 2022

Given below are two statements : One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): The stretching of a spring is determined by the shear modulus of the material of the spring. Reason (R): A coil spring of copper has more tensile strength than a steel spring of same dimensions. In the light of the above statements, choose the most appropriate answer from the options given below

1(A) is false but (R) is true

2Both (A) and (R) are true and (R) is the correct explanation of (A)

3Both (A) and (R) are true and (R) is not the correct explanation of (A)

4(A) is true but (R) is false

NTA Answer: Option 4(final)

NEET 2021

The velocity of a small ball of mass M and density d, when dropped in a container filled with glycerine becomes constant after some time. If the density of d glycerine is , then the viscous force acting on the 2 ball will be Mg

12Mg

22 3 Mg

3Mg

NTA Answer: Option 2(final)

NEET 2021

A cup coffee cools from 90°C to 80°C in t minutes, when the room temperature is 20°C. The time taken by a similar cup of coffee to cool from 80°C to 60°C at a room temperature same at 20°C is 5 13 t t

1(2) 13 10 13 10 t t

2d hc   2m c  λ2 λ =  

3(4) 5 13

4d h  

NTA Answer: Option 3(final)

NEET 2020

r U U 17. A capillary tube of radius r is immersed in water ° h – ÷ U and water rises in it to a height h. The mass of the water in the capillary is 5 g. Another capillary 5 g – 2r tube of radius 2r is immersed in water. The mass U – U — of water that will rise in this tube is :

22.5 g

35.0 g

NTA Answer: Option 4(final)

NEET 2020

r U r (r =1.5 r ) U U 45. The quantities of heat required to raise the 1 2 1 2 1 K fl h U fl c temperature of two solid copper spheres of radii — r and r (r =1.5 r ) through 1 K are in the 1 2 1 2 5 ratio :

127 3

29 8

33 4

NTA Answer: Option 2(final)

How NEET usually asks this

5 recurring patterns from past papers — click to collapse

Apply Bernoulli's equation to flow through pipes of varying cross-section / heights / Venturi-like geometries.

InterpretationMedium

Common distractors

forgets height term

Drops rho*g*h term

equates pressures incorrectly

Picks wrong reference points

Two bodies of given material and dimensions; find ratio of heat needed for same temp change. Q ∝ m c.

Direct ApplicationEasy

Common distractors

ignores mass difference

Compares only specific heats, not masses

Energy to form a bubble of given radius from soap solution; or pressure inside vs outside. Bubble has 2 surfaces (factor 4T/r).

Direct ApplicationEasy

Common distractors

uses 2T r instead of 4T r for bubble

Treats soap bubble like a drop

Sphere falling in viscous liquid; find terminal velocity or shape of v(t) curve. v_t ∝ r^2.

InterpretationMedium

Common distractors

expects linear acceleration

Default to constant acceleration without recognising drag-induced terminal v

Wire stretching: given Y, length, area, force or stress, find elongation or stress at limit. Y = FL/(A delta_L).

Multi StepMedium

Common distractors

uses bulk modulus formula

Confuses Y with K

Sources

NCERT refs: Class 11 Physics Chapter 11, p.7

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Capillary Rise Effect of Gravity on Pressure

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Change of State Latent Heat

Lesson

Practice MCQs

Quick recall before you leave

Latent heat formula

Specific heat formula

Worked Example

Before solving, remember these

Formulas

Valid when

Valid when

Valid when

Valid when

Valid when

Valid when

Valid when

Valid when

Valid when

Valid when

Valid when

Valid when

Exam Traps & Common Mistakes

When it triggers

How to avoid

When it triggers

How to avoid

Correction

Correction

Correction

Past Year Questions

How NEET usually asks this

Common distractors

Common distractors

Common distractors

Common distractors

Common distractors

Sources

Free NEET study resources